Optocoupler question

Discussion in 'The Projects Forum' started by dswartz2, Jun 4, 2009.

  1. dswartz2

    Thread Starter Active Member

    May 19, 2009
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    0
    Hi all.

    I am using a dual channel ILD1 optocoupler, phototransistor output.

    My input voltage is 48 Vdc. I have a 2.4 kohm resistor connected at the input to get the desired forward current of 20mA.
    At the output of the opto, I have a 2.2 kohm resistor hooked up to my desired output voltage of 24 Vdc.

    The forward voltage on the data sheet is typically 1.25V with a max of 1.65V. When I measure the voltage going in to the opto, it is about 1.22V. The voltage at the output before the 24V source, is around 0.3V.

    I want to know what I need to do to get the transister to fully turn on and get my output to be my 24Vdc. If I need to get my forward voltage higher, how should I go about doing that?

    Thank you very much in advance for any help!
     
  2. franzschluter

    Active Member

    Jun 1, 2009
    95
    0
    I think you just fried your optocoupler... Input voltage of typical Optocouplers is TTL... 6V maximum! Even 48V at the output is barely approaching critical mark of absolute rating dude..

    To fix your circuit.. You can perhaps create a voltage divider network.. Such that your 48V will be divided by two Resistors so you will get a smaller voltage from the big chunk of 48V.. Then place your input in between the resistor and reference such that you have a potential difference of 4.8V.. Next is making your "Rb" or input resistance.. The current flowing to your input must not exceed 60mA.. So playing with 30~40mA is more safe... Typical Rb will be 100 Ohms in series to limit input current. Then at your output check your voltage which is 24V.. ILD1 can take 400mA.. So compute now how much resistance you must need to limit max 400mA and not exceed this range.. Or add the loads in your equation if you have coils or capacitors as load.

    Regards
    Franz
     
  3. dswartz2

    Thread Starter Active Member

    May 19, 2009
    31
    0
    ok...

    So I designed a voltage divider. R1 = 2160 ohm and R2 = 240 ohm. This gives me 4.8V.

    Now, do I need to add another resistor in between the 4.8V and the input of the opto? If so, another 240 ohm resistor would give me 20mA in to the input. Does that sound right? or do I not need the extra resistor?
     
  4. franzschluter

    Active Member

    Jun 1, 2009
    95
    0
    Yes thats ok :D.. Try using multisim if you are as lazy as me to compute for current and resistor values etc etc etc.
     
  5. dswartz2

    Thread Starter Active Member

    May 19, 2009
    31
    0
    alright thanks.

    I'll just run through what I have real quick just to make sure that everything is right before I mess this up again... if you could, let me know if it sounds good.


    So I have 48Vdc hooked up to a voltage divider that gives out 4.8V. Connected to the 4.8V, I have a 240 ohm resistor that will limit the current to 20mA going in to the opto. On the collector part of the output of the opto, I have a 24V source connected with a resistor that limits a 400mA max. Also connected to the collector is my load. And the emitter output of the opto is connected to ground.

    does this sound correct?
     
  6. dswartz2

    Thread Starter Active Member

    May 19, 2009
    31
    0
    Now that I think about it... the emitter output should be connected to my load shouldn't it?

    Because when the opto turns on, it will then connect the 24 volts to my load.
     
  7. franzschluter

    Active Member

    Jun 1, 2009
    95
    0
    Other way around dude...

    There are many types of Optocouplers dude. Essentially imagine it as there is a transistor inside it...
    So if its collector output.. This means its somewhat like a NPN...
    To make it more clear...

    Example: pin 1 is anode and 2 is cathode.. You connect your 4.8V+ to your anode.. then your ground in the cathode...
    Once there is current flowing through those two terminals.. An internal LED will light up inside and switch the internal photo transistor "on".. On the other side at Pins 8 and 7 will start to conduct.. Now take note if its NPN style or PNP..


    NPN style:
    You connect your load on the high-side. Meaning.. Collector goes to the 1pin of your load.. the other pin of your load goes to the positive of your +24V.. Now for emitter you simply connect that to ground.. That is also called high-side switching.. Never ever put negative in the collector of an NPN phototransistor,N-channel MOS,NPN BJT or any of these... As they will forward conduct and burn or make a good water heater..So check polarity..

    You almost did it right.. But you are doing low-side switching..Also known as emitter follower circuit.. This will not work in NPN only in PNP. It might or might not work, better is do high-side switch..Don't worry if you did this, it will not burn.. But now try putting load on the collector not emitter...When you read datasheet you will see what I mean...

    http://www.semiconductors.com.pl/web/pliki/ild5.pdf
    Look carefully at Non saturated switching and saturated switching.. What you want is saturated switching ;)

    PNP style:
    Works exactly like the NPN except you connect your load to the low side (Load is in between ground and emitter).. Also take note your collector is now negative or ground and emitter positive.. And it'll turn on when input is low.. and turn off when input is high..

    So let's assume you burned 1 input/output unit.. No worries.. ILD1 usually has two units inside it.. You may still use pins 3 and 4 and output 6 and 8..

    Regards
    Franz
     
    Last edited: Jun 4, 2009
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